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It is well-known that all permutations can be written as a product of 2-cycles. Furthermore, it is relatively trivial to show that all even permutations can be written as a product of 3-cycles (but odd permutations not [this would break parity]). However, is there a nice way to go about this for prime $p$-cycles? (I.e. what are the conditions for a permutation to be written as a product of $p$-cycles?)

I tried to find information online, but google gives me information about topics with close keywords, such as permutations of order $p$.

By simple parity analysis, one can say that products of $p$-cycles are even (for $p>2$). Furthermore, if a permutation cannot be written as a product of a multiple of $p-1$ $2$-cycles (i.e. the product is of length $n(p-1)$), then you cannot represent it as a product of $p$-cycles (since $(12\dots p)=(1p)(1(p-1))\dots (12)$), but I am not sure if this is a sufficient condition, since my method for tackling the $3$-cycle case was already quite brute force (show that all possible products of $2$-cycles can be written as $3$-cycles), so it is not generalisable.

Anyways, if anyone has any information on this matter (even just giving me the right keywords to research the topic further), I would be glad for any help.

110112345
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    the subgroup generated by all $p$-cycles is normal, so for $n\geq5$ it is either $S_n$ or $A_n$ (or just $1$ if $p>n$). Since $p$-cycles are even for $p>2$, it will be $A_n$. – user8268 Sep 10 '24 at 12:57
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    @user8268 I could not find a duplicate. Yours is a more elegant solution than mine. Consider posting it as answer (unless it's a dupe). If you find a duplicate, do let me know, please. Gotta commute next. – Jyrki Lahtonen Sep 10 '24 at 13:05
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    Please do not answer questions in the comments, @user8268. – Shaun Sep 10 '24 at 13:20
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    @Shaun you are right, I made it to an answer – user8268 Sep 10 '24 at 13:59

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Hint: $$(1234567)(7654312)=(132).$$

Jyrki Lahtonen
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  • See this thread for more interesting results. – Jyrki Lahtonen Sep 10 '24 at 12:58
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    Your hint shows this has little to do with the cycle lengths being a common odd prime number, other than the question being in that context. They could be any common odd number that is greater than or equal to $3$ with the same result. – KCd Sep 10 '24 at 13:00
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    Very true, @KCd. Not much to this. An exercise for a first course on groups. – Jyrki Lahtonen Sep 10 '24 at 13:03
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    I will happily delete this, when the OP can post a complete solution. Not too much left to do. – Jyrki Lahtonen Sep 10 '24 at 13:04
  • Thank you. Yes indeed this was motivated by an exercise in a first course on groups (we were asked to consider the 3-cycle case, but I immediately thought of $p$-cycles, as I had an intuition that they would have something nice, since prime numbers are important have nice properties with finite groups). I should have probably realised on my own that it being prime or odd is irrelevant. But again thank you. – 110112345 Sep 10 '24 at 13:35
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    @110112345 The length being odd is relevant, since that makes the cycles of that length even and thus each one is in $A_n$. – KCd Sep 10 '24 at 13:40
  • @KCd I meant that it was irrelevant that it was particularly a prime-cycle since it works for odd-cycles as well. (Which was something I didn't realise while posing this question for myself). – 110112345 Sep 10 '24 at 13:41
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(moved from the comments)

The subgroup generated by all $p$-cycles is normal in $S_n$, so for $n\geq5$ it must be $A_n$ or $S_n$ (or $1$ if $p>n$). Since $p$-cycles are even for $p\neq2$, it is $A_n$.

user8268
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Alright, thank you for the hint @JyrkiLahtonen

So, I was close. We already know that no odd permutation can be written as a product of $p$-cycles. But now as clearly for odd $p$ (also prime $p>2$), we can write $(132)=(123\dots p)(p(p-1)\dots 312)$, so all $3$-cycles can be written as a product of $p$-cycles. But we already know that all even permutations can be written as a product of $3$-cycles, so they can be written as a product of $p$-cycles as well.

Therefore the answer to my question becomes that the in order for a permutation to be written as a product of $p$-cycles the permutation has to be even.

110112345
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  • Correct. You can accept your own answer, I think. After all, mine is just a hint. I would have posted it as a comment, but some others criticize the practice :-) – Jyrki Lahtonen Sep 10 '24 at 13:38
  • @JyrkiLahtonen I could, but it says "accept this answer if it solved your problem or was the most helpful in finding your solution" and I think your answer satisfies that criterion. Although, I am unsure of exact practices here. – 110112345 Sep 10 '24 at 13:40
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Proposition Let $n$ be a natural number and $1 \lt k \leq n$. Let $H_k$ be the subgroup of $S_n$ generated by all the $k$-cycles. Then $H_k=S_n$ if $k$ is even, and $H_k=A_n$ if $k$ is odd.

Note For $k=2$ this is the well-known fact that $S_n$ is generated by its transpositions. For $k=3$ this yields another well-known fact that $A_n$ is generated by $3$-cycles. The latter is used in the proof.

Proof We can assume that $k \geq 3$. Let $\{a_1,a_2, \ldots , a_k\} \subseteq \{1,2, \dots, n-1, n \}$, with all the $a$'s different. Then note that the product of $k$-cycles $$(a_k a_{k-1} \cdots a_2 a_1)(a_1 a_3 a_2 a_4 \cdots a_{k-1} a_k)=(a_1 a_2 a_3)$$ Hence, with $k$-cycles one can generate any $3$-cycle. So $A_n \subseteq H_k \subseteq S_n$. If $k$ is odd then $H_k$ only contains even permutations and hence $H_k=A_n$. If $k$ is even, then the $k$-cycles in $H_k$ are odd permutations and since $|S_n:A_n|=2$, we must have $H_k=S_n$. $\square$

Nicky Hekster
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